An electromagnet has stored 540 J of magnetic energy when a current of 6 A exists in its coil. What is the value of average emf induced, if the current is reduced to zero in 0.54 s?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 333.3 V
The magnetic energy stored in a coil of self inductance L carrying current I is \( U = \frac{1}{2}LI^{2} \), and the average emf induced when the current changes by \( \Delta I \) in time \( \Delta t \) is \( \varepsilon = L\,\frac{|\Delta I|}{\Delta t} \). First find L from the stored energy.
\( \displaystyle L = \frac{2U}{I^{2}} = \frac{2\times540\,\mathrm{J}}{(6\,\mathrm{A})^{2}} = 30\,\mathrm{H} \)
The current falls from 6 A to 0, so \( |\Delta I| = 6\,\mathrm{A} \), and \( \Delta t = 0.54\,\mathrm{s} \).
\( \displaystyle \varepsilon = L\,\frac{|\Delta I|}{\Delta t} = 30\,\mathrm{H}\times\frac{6\,\mathrm{A}}{0.54\,\mathrm{s}} = 333.3\,\mathrm{V} \)
Shortcut: Eliminating L between the two relations gives \( \varepsilon = \frac{2U}{I\,\Delta t} \):
\( \displaystyle \varepsilon = \frac{2\times540}{6\times0.54} = 333.3\,\mathrm{V} \)